Fourier's Law (Heat Conduction)

Find out how much heat passes through a wall, panel or plate of a given material and thickness when you know the temperature difference across it.

How the calculation works

Fourier's law for steady one-dimensional conduction through a flat, uniform slab gives q = k·ΔT / d and Q = q·A. Here k (λ) is the material's thermal conductivity in W/(m·K), A is the area in m², ΔT is the temperature difference between the two faces in kelvin (a 1 K difference equals a 1 °C difference), and d is the thickness in meters.

The main result is the heat flow Q in watts. You also get the heat flux q in W/m², the layer's thermal resistance R = d / k in m²·K/W (the metric RSI value; multiply by about 5.678 to get the US R-value in ft²·°F·h/Btu), and the energy lost in 24 hours, Q × 24 / 1000, in kWh. A negative ΔT simply reverses the direction of flow.

Typical conductivities: mineral wool about 0.035–0.045, softwood across the grain 0.12–0.18, solid brick masonry 0.5–0.8, reinforced concrete about 1.7–2.0, steel about 50 and copper about 390 W/(m·K). Use the manufacturer's declared value for design work.

Worked example

Defaults: k = 0.5 W/(m·K), area 10 m², ΔT = 20 K, thickness 0.1 m. Heat flux q = 0.5 × 20 / 0.1 = 100 W/m², heat flow Q = 100 × 10 = 1,000 W. Thermal resistance R = 0.1 / 0.5 = 0.2 m²·K/W (about R-1.1 in US units), and the daily loss is 1,000 × 24 / 1,000 = 24 kWh.

Things to keep in mind

• ΔT here is the difference between the wall's surfaces, not the indoor and outdoor air. To work from air temperatures, add surface resistances; ISO 6946 typically uses Rsi = 0.13 and Rse = 0.04 m²·K/W.
• For a layered wall add the resistances of the layers: q = ΔT / (R₁ + R₂ + …). This calculator handles a single uniform layer.
• Damp insulation conducts heat much better than dry material, so a catalogue k for dry insulation gives an optimistic estimate.
• The model is steady-state: it ignores heat storage in massive walls and thermal bridges through studs, fixings and joints.